Question

there given a trapezium with AB parallel DC , proove angle A =angle B
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Answer 1

Step-by-step explanation:

We have AB || CD and AD = BC

(i) To prove that ∠A = ∠B.

Produce AB to E and draw CE || AD.

∴ AB || DC

⇒ AE || DC   [Given]

Also AD || CE  [Construction]

∴ AECD is a parallelogram.

⇒ AD = CE   [opposite sides of the parallelogram AECD]

But AD = BC  [Given]

∴ BC = CE

Now, in ΔBCE, we have

BC = CE

⇒ ∠CBE = ∠CEB    …(1)

[∵ Angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180°  [Linear pair] ...(2)

and ∠A + ∠CEB = 180°  [∵ Adjacent angles of a parallelogram are supplementary] …(3)

From (2) and (3), we get

∠ABC + ∠CBE = ∠A + ∠CEB

But ∠CBE = ∠CEB    [Using (1)]

∴ ∠ABC = ∠A

or ∠B= ∠A

or ∠A= ∠B

(ii) To prove that ∠C = ∠D.

AB || CD and AD is a transversal.

∴ ∠A + ∠D = 180º   [Sum of interior opposite angles]

Similarly, ∠B + ∠C = 180º

⇒ ∠A + ∠D= ∠B + ∠C

But ∠A= ∠B  [Proved]

∴ ∠C= ∠D

(iii) To prove ΔABC ≌ ΔBAD

In ΔABC and ΔBAD, we have



AB = BA   [Common]

BC = AD   [Given]

∠ABC = ∠BAD    [Proved]

∴ ΔABC ≌ ΔBAD  [Using SAS criteria]

(iv) To prove that diagonal AC = diagonal BD ∵

ΔABC ≌ ΔBAD   [Proved]

∴ Their corresponding parts are equal.

⇒ the diagonal AC = the diagonal BD.

Hope it was helpful